1 ˆ

Ψ(C) = µ(trC ’ 3) (5.49)

2

Now using Equation (5.46) S is obtained with the help of Equations (5.19a)

and (5.20) as,

‚ Ψ(C)

+ pJC ’1

S=2

‚C

ˆ

‚trC

+ pJC ’1

=µ

‚C

‚ ’1/3

(IIIC C : I) + pJC ’1

=µ

‚C

’1/3 ’1/3’1

I ’ 1 IIIC IIIC C ’1 (C : I)] + pJC ’1

= µ[IIIC 3

’1/3

(I ’ 1 IC C ’1 ) + pJC ’1

= µIIIC (5.50)

3

The corresponding Cauchy stress tensor can now be obtained by using

Equation (4.45b) to give σ as,

σ = J ’1 F SF T

= µJ ’5/3 F (I ’ 1 IC C ’1 )F T + pF C ’1 F T

3

1

σ = µJ ’5/3 (b ’ 3 IbI)

= σ + pI ; (5.51)

where the fact that Ib = IC has been used again.

15

5.5 INCOMPRESSIBLE MATERIALS

We can now evaluate the Lagrangian elasticity tensor with the help

of Equations (5.10) or (5.11). The result can be split into deviatoric and

pressure components, C and C p respectively, as,

‚2Ψ ‚(JC ’1 )

‚S ‚S

C=2 = C + Cp; C=2 =4 ; C p = 2p

‚C ‚C ‚C‚C ‚C

(5.52)

With the help of Equations (5.22) and (5.31) these two components can be

evaluated for the neo-Hookean case de¬ned by Equation (5.49) after lengthy

but simple algebra as,

’1/3 1

’ 1 I — C ’1 ’ 3 C ’1 — I + 9 IC C ’1 — C ’1

1 1

C = 2µIIIC 3 IC III (5.53a)

3

C p = pJ[C ’1 — C ’1 ’ 2I ] (5.53b)

Note that the pressure component C p does not depend on the particular

material de¬nition being used.

The spatial elasticity tensor is obtained by the push forward type of

operation shown in Equation (5.14) as,

c = J ’1 φ— [C]; c p = J ’1 φ— [C p ]

ˆ ˆ

c = c +c p ; (5.54)

Performing this push forward operation in Equations (5.53a,b) gives,

c = 2µJ ’5/3 1 ICi ’ 3 b — I ’ 1 I — b + 1 IbI — I

1

(5.55a)

3 3 9

cp = p[I — I ’ 2i ] (5.55b)

EXAMPLE 5.5: Mooney“Rivlin materials

A general form for the strain energy function of incompressible rubbers

attributable to Mooney and Rivlin is expressed as,

µrs (IC ’ 3)r (IIC ’ 3)s

—

Ψ(C) =

r,s≥0

—

where IIC is the second invariant of C de¬ned as,

2

IIC = 1 (IC ’ IIC );

—

IIC = C : C

2

The most frequently used of this family of equations is obtained when

only µ01 and µ10 are di¬erent from zero. In this particular case we

have,

2

1

Ψ(C) = µ10 (IC ’ 3) + 2 µ01 (IC ’ IIC ’ 6)

16 HYPERELASTICITY

EXAMPLE 5.5 (cont.)

The equivalent homogeneous potential is obtained by replacing C by

ˆ

C in this equation to give,

ˆ ˆ ˆˆ

Ψ(C) = µ10 (trC ’ 3) + 2 µ01 [(trC)2 ’ C : C ’ 6]

1

5.5.3 NEARLY INCOMPRESSIBLE HYPERELASTIC

MATERIALS

As explained at the beginning of Section 5.5 near incompressibility is often

a device by which incompressibility can more readily be enforced within

the context of the ¬nite element formulation. This is facilitated by adding a

volumetric energy component U (J) to the distortional component Ψ already

de¬ned to give the total strain energy function Ψ(C) as,

Ψ(C) = Ψ(C) + U (J) (5.56)

where the simplest example of a volumetric function U (J) is,

U (J) = 2 κ(J ’ 1)2

1

(5.57)

It will be seen in Chapter 6 that when equilibrium is expressed in a varia-

tional framework, the use of Equation (5.57) with a large so-called penalty

number κ will approximately enforce incompressibility. Typically, values of

κ in the region of 103 ’104 µ are used for this purpose. Nevertheless, we must

emphasize that κ can represent a true material property, namely the bulk

modulus, for a compressible material that happens to have a hyperelastic

strain energy function in the form given by Equations (5.56) and (5.57).

The second Piola“Kirchho¬ tensor for a material de¬ned by (5.56) is

obtained in the standard manner with the help of Equation (5.22) and noting

that IIIC = J 2 to give,

‚Ψ

S=2

‚C

‚Ψ dU ‚J

=2 +2

‚C dJ ‚C

‚Ψ

+ pJC ’1

=2 (5.58)

‚C

where, by comparison with (5.46), we have identi¬ed the pressure as,

dU

p= (5.59)

dJ

17

5.5 INCOMPRESSIBLE MATERIALS

which for the case where U (J) is given by Equation (5.57) gives,

p = k(J ’ 1) (5.60)

This value of the pressure can be substituted into the general Equation (5.58)

or into the particular Equation (5.50) for the neo-Hookean case to yield the

complete second Piola“Kirchho¬ tensor. Alternatively, in the neo-Hookean

case, p can be substituted into Equation (5.51) to give the Cauchy stress

tensor.

EXAMPLE 5.6: Simple shear (ii)

Again we can study the case of simple shear for a nearly incompressible

neo-Hookean material. Using Equation (5.51) and the b tensor given

in Exercise 5.4 we obtain,

®2 2

γ γ 0

3

1

’ 3 γ2

σ = µ° γ 0»

’ 1 γ2

0 0 3

where now the pressure is zero as J = 1 for this type of deformation.

Note that for this type of material there is no Kelvin e¬ect in the sense

that a volume-preserving motion leads to a purely deviatoric stress

tensor.

EXAMPLE 5.7: Pure dilatation (ii)

It is also useful to examine the consequences of a pure dilatation on a

nearly incompressible material. Recalling that this type of deformation

has an associated left Cauchy“Green tensor b = J 2/3 I whose trace is

Ib = 3J 2/3 , Equations (5.51) and (5.60) give,

σ = κ(J ’ 1)I

As expected a purely dilatational deformation leads to a hydrostatic

state of stresses. Note also that the isochoric potential Ψ plays no role

in the value of the pressure p.

Again, to complete the description of this type of material it is necessary

to derive the Lagrangian and spatial elasticity tensors. The Lagrangian

tensor can be split into three components given as,

‚(JC ’1 )

‚Ψ ‚p

+ 2JC ’1 —

C=4 + 2p (5.61)

‚C‚C ‚C ‚C

18 HYPERELASTICITY

The ¬rst two components in this expression are C and Cp as evaluated in

the previous section in Equations (5.53a,b). The ¬nal term, namely C κ , rep-

resents a volumetric tangent component and follows from U (J) and Equa-

tion (5.22) as,

‚p

C κ = 2JC ’1 —

‚C

d2 U ’1 ‚J

= 2J C—

dJ 2 ‚C

d2 U ’1

2

C — C ’1

=J (5.62)

2

dJ

which in the case U (J) = κ(J ’ 1)2 /2 becomes,

C κ = κJ 2 C ’1 — C ’1 (5.63)

Finally, the spatial elasticity tensor is obtained by standard push forward

operation to yield,

c = J ’1 φ— [C] = c + cp +c κ

ˆ (5.64)

ˆ

where the deviatoric and pressure components, c and cp respectively, are

identical to those derived in the previous section and the volumetric com-

ponent c κ is,

d2 U

’1

cκ = J φ— [C κ ] = J 2 I — I (5.65)

dJ

which for the particular function U (J) de¬ned in Equation (5.57) gives,

c κ = κJ I — I (5.66)

Remark 5.5.1. At the initial con¬guration, F = C = b = I, J = 1,

p = 0, and the above elasticity tensor becomes,

ˆ

c = c +c κ

= 2µ i ’ 1 I — I + κI — I

3

= κ ’ 2 µ I — I + 2µi (5.67)

3

which coincides with the standard spatially isotropic elasticity tensor (5.37)

with the relationship between » and κ given as,

2

» = κ ’ 3µ (5.68)

In fact all isotropic hyperelastic materials have initial elasticity tensors as

de¬ned by Equation (5.37).

19

5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS

5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS

5.6.1 MATERIAL DESCRIPTION

It is often the case that the constitutive equations of a material are presented

in terms of the stretches »1 , »2 , »3 in the principal directions N 1 , N 2 , and

N 3 as de¬ned in Section 3.5. In the case of hyperelasticity, this assumes that

the stored elastic energy function is obtainable in terms of »± rather than

the invariants of C. This is most likely to be the case in the experimental